Solving Problems Algebraically

Solving Problems Algebraically-60
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The equation must first of all be written in standard form, and then the coefficients plugged into the formula.

The formula was derived in class by completing the square on a generic quadratic equation.

how many solutions does the following system of linear equations have i have my system right over here there's a couple of ways to think about it one way is to think about them graphically and think about well, are they the same line in which case they would have an infinite number of solutions or they are parallel in which case they never intersect you would have no solution or they intersect exactly at one place, in that case, you would have exactly one solution but instead, we're going to do this algebraically So let's try to actually just solve the system and see what we get So, the first equation...

I will leave that unchanged 5x-9y=16 Now this second equation right over here, let's say I wanna cancel out the x terms so let me multiply the second equation by negative 1 so I have a -5x that I can cancel out with the 5x so if I multiply this second equation by -1 we will have -5x 9y = -36 Now I'm going to add to the left side of the equation and the right side of the equation to get a new equation so 5x - 5x well that's going to make a zero -9y 9y well that's gonna be zero again i don't even have to write it, it's gonna be zero on the left side and on the right-hand side I'm gonna have 16 - 36 = -20 So now I'm left with the somewhat bizarre looking equation that says that 0 is equal to -20 Now one way, you might say, "Well-well how does this make any sense?

Works well when there is no linear term, that is, when B=0.

The extraction of roots is called the square root principle by your text.

The goal here is to get the squared variable term by itself on one side and a non-negative constant on the other side. Remember that the square root of x is the absolute value of x.

When you solve an equation involving an absolute value, you will get a plus and minus in the solution.

Works well when the quadratic can easily be factored.

The AC Method is explained elsewhere if you want to review or learn it.


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